ap calc princetown review example 4

Test Preparation AP Test Preparation
1

<p>ok, so I thought I would review some calc from the 2006-2007 princetown review, but im confused about example 4 on pg 196. </p>

<p>How is it that when you integrate y=sqrt(3-x) from 0 to 3, the area is negative? You get -2sqrt(3).</p>

<p>I thought that areas were always positive above the x axis. Can someone please clarify? Thanks.</p>

2

<p>I get 2sqrt(3).</p>

<p>u = 3 - x
du = -dx
Int( -u^.5 du) = -(2/3)u^1.5 = -(2/3)*(3-x)^1.5</p>

<p>Integrate from 0 to 3
-(2/3)(3-3)^1.5 - (-2/3)(3-0)^1.5
= 0 + (2/3) * 3 * sqrt(3)
= 2sqrt(3)</p>

3

<p>!!!! Thanks. I integrated without using substitution lol.</p>